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IrinaVladis [17]
3 years ago
15

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 42 g of water (with an

initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal?
Chemistry
1 answer:
kogti [31]3 years ago
8 0

The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

-Qm = Qw

Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

where C is the specific heat capacities of the materials.

We calculate as follows:

-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)

-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)

Cm = 1.717 -----> OPTION C

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abruzzese [7]

Answer:

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Explanation:

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3 years ago
Chlorine has two naturally occurring isotopes, ³⁵Cl and ³⁷Cl. What is the mass number of each isotope? How many protons, neutron
Arte-miy333 [17]

The mass number of ³⁵Cl and ³⁷Cl is 35 and respectively.

<h3>What is atomic mass? </h3>

Atomic mass is defined as the sum of number of protons and number of neutrons of an element.

Atomic mass is represented by A.

<h3>What is Atomic number?</h3>

Atomic number is defined as the number of electron in an element.

Atomic number is represented by Z.

<h3>What is Isotopes? </h3>

Isotopes are the element which have same atomic number but have different atomic mass.

Number of proton is equal to number of electrons.

Proton is positively charged, electrons are negativity charge while neutrons are neutral in nature.

Proton and neutrons are present in the nucleus of an element. On the other hand, electron revolves around the nucleus.

The atomic number of ³⁵Cl and ³⁷Cl is 17.

Number of protons = Number of electrons = 18.

Neutrons = atomic mass - atomic number

Neutrons of ³⁵Cl = 35-17 = 18

Neutrons of ³⁷Cl= 37- 17= 20

So, they have different atomic mass and number of neutrons.

Thus, we concluded that the mass number of ³⁵Cl and ³⁷Cl is 35 and 37 respectively.

learn more about atomic mass:

brainly.com/question/14250653

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6 0
2 years ago
Caluclate the volume of0,25 mol dm -^3 hydrochloric acid required to neutralize 2.0 dm^3 of 0,15 mol dm-^3 barium hydroxide
inessss [21]

Answer:

2.4 dm-3

Explanation:

Equation of the reaction;

2HCl(aq) + Ba(OH)2(aq) -------> BaCl2(aq) + 2 H2O(l)

Volume of acid VA = ??

Concentration of acid CA = 0.25 moldm-3

Volume of base VB = 2.0 dm^3

Concentration of base CB = 0.15 moldm-3

Number of moles of acid NA = 2

Number of moles of base NB = 1

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VA = CBVBNA/CANB

VA = 0.15 * 2 * 2/0.25 *1

VA = 2.4 dm-3

4 0
3 years ago
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Afina-wow [57]

Answer:

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Explanation:

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3 years ago
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An unknown compound was found to have a percent composition as follows: 47.0%, 14.5 carbon, and 38.5 oxygen. What is its empiric
shusha [124]

 KCO₂

 K₂C₂O₄

Explanation:

Given parameters:

Percent composition:

             K = 47%

             C = 14.5%

             O = 38.5%

Molar mass of compound = 166.22g/mol

Unknown:

Empirical formula of compound = ?

Molecular formula of compound = ?

Solution:

The empirical formula of a compound is its simplest formula. Here is how to solve for it:

 

                                K                                C                           O

Percent

composition           47                               14.5                       38.5

Molar mass            39                                  12                           16

Number

of moles               47/39                         14.5/12                     38.5/16

moles                    1.205                          1.208                          2.4

Dividing

by smallest      1.205/1.205               1.208/1.205                      2.4/1.205

                                1                                  1                                        2

Empirical formula               KCO₂

Molecular formula

  This is the actual combination of the atoms:

          Molecular formula =   ( empirical formula of KCO₂)ₙ

 Molar mass of empirical formula = 39 + 12 + 2(16) = 83g/mol

      n factor = \frac{true molecular mass}{molar mas of empirical formula}

      n factor = \frac{166.22}{83} =  2

Molecular formula of compound = ( KCO₂)₂ = K₂C₂O₄

Learn more:

Empirical formula brainly.com/question/2790794

#learnwithBrainly

6 0
4 years ago
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